TITLE: Proof of ${F(n+4)}^{4} - {4F(n+3)}^{4} - {19F(n+2)}^{4} - {4F(n+1)}^{4}+{F(n)}^{4} = -6$
QUESTION [9 upvotes]: Observe:
\begin{matrix} F(n)|&{F(n)}^{4}& - {4F(n+1)}^{4}& - {19F(n+2)}^{4}&- {4F(n+3)}^{4}&{F(n+4)}^{4}& = -6\\ 1|& 1& -4& -304& -324& 625&=-6\\ 1|& 1& -64& -1539& -2500& 4096&=-6\\ 2|& 16& -324& -11875& -16384& 28561&=-6\\ 3|& 81& -2500& -77824& -114244& 194481&=-6\\ 5|& 625& -16384& -542659& -777924& 1336336&=-6\\ 8|& 4096& -114244& -3695139& -5345344& 9150625&=-6 \end{matrix}
I see that the proof is true but I cant quite grasp the pattern.  I'm more interested in hints rather than a solution. Ive read through this website about Fibonomials a couple times and have a decent understanding but I cant figure how to apply it in this situation. It may not even be necessary. I've tried proving through induction  but ended up with almost exactly the same problem. I think there may be a proof through some sort of recursive sequence but I don't know enough to prove something like that.
I'm more interested in hints rather than a solution.

REPLY [3 votes]: You already have an accepted answer, but here is an alternative approach.
$${F(n+4)}^{4} - {4F(n+3)}^{4} - {19F(n+2)}^{4} - {4F(n+1)}^{4}+{F(n)}^{4}$$
reduces to
$$-6\left(F(n+1)^2 - F(n+1)F(n) - F(n)^2\right)^2$$ 
by applying Fibonacci recursion.
So what's left to prove is that
$$F(n+1)^2 - F(n+1)F(n) - F(n)^2 \in \{1, -1\}$$
or more specifically:
$$F(n+1)^2 - F(n+1)F(n) - F(n)^2 = \begin{cases} 
+1 \text{ for } 2\mid n \\
-1 \text{ for } 2\not\mid n \\
\end{cases}$$
which follows inductively, given
$$\begin{align}
&  F(n+1)^2 - F(n+1)F(n) - F(n)^2\\
 &= F(n+1)\bigg(F(n+1) - F(n)\bigg) - F(n)^2 \\ 
 &= \bigg(F(n) + F(n-1)\bigg)F(n-1) - F(n)^2 \\
 &= -\left(F(n)^2 - F(n)F(n-1) - F(n-1)^2\right)
\end{align}$$